Let's Get This Down On Paper: September 2015

A pendulum swinging with a small amplitude is a good approximation of simple harmonic motion in the horizontal direction. What path would a pendulum bob bake in order for the horizontal motion to be simple harmonic motion?

A pendulum is driven by two forces, gravity pulling the mass downward and tension pulling the mass toward the string. The result is a total net force, mg

\cdot

sin(θ), that is oriented perpendicular to the string:

http://www.webassign.net/question_assets/asucolphysmechl1/lab_7/images/figure7-1-alt.png

What I'm interested in is not the mass being confined to a spring, but a mass being confined to a path, with a gravitational field present. So, let's consider a frictionless point particle on a bowl, which at any instant looks like a mass on an incline:

http://www.webassign.net/question_assets/tamucolphysmechl1/lab_2/images/figure2-1.png

We want simple harmonic motion in the x-direction:

$F(x) = -k\cdot sin(\Theta )$

We also know by studying the above diagram that the in the x-component of the net force will only depend on the normal force since gravity is always in the y-direction. The value of x-component of force is:

$Fx(\Theta ) = T\cdot sin(\Theta )$

Tension is variable and will equal the component of gravity that is along the direction of the string:

$T(\Theta )= mg\cdot cos(\Theta )$

Since the x-component of the tension is the only force present in the horizontal direction, we can set it equal to our simple harmonic motion constraint:

$-kx = mg\cdot cos(\Theta )sin(\Theta )$

*** Here is where the magic happens ***

What is the actual meaning of the angle? Well, for a pendulum it is the angle between the string and vertical, but for a mass on an incline it is the incline angle. If the mass were on a smooth incline we could replace the trigonometric terms in our latest equation with the ratios they represent. But-- actually-- we can still do that! Here's how it works:

This triangle represents the instantaneous slope at any particular point on the ball's path. The triangle was scaled to have a base of 1, which allows the side to have a component equal to the slope of the function. Think of it this way-- slope is "rise over run", slope is also dy/dx. I'm just setting dy/dx = (dy/dx)/1. The hypotenuse can then be found using the Pythagorean Theorem.

Can now see the following relationships:

$sin(\Theta )= \frac{\frac{dy}{dx}}{(1 + (\frac{dy}{dx})^2)^\frac{1}{2}}$

$cos(\Theta )= \frac{1}{(1 + (\frac{dy}{dx})^2)^\frac{1}{2}}$

Combining these two with our equation before the magic happened:

$-kx = mg \cdot \frac{1}{(1+ \frac{dy}{dx})^{\frac{1}{2}}} \cdot \frac{\frac{dy}{dx}}{(1+ \frac{dy}{dx})^{\frac{1}{2}}}$

Simplifying:

$-kx = mg \cdot \frac{\frac{dy}{dx}}{1+ (\frac{dy}{dx})^{2}}$

Now let's rearrange this differential equation:

$x + x(\frac{dy}{dx})^2 + \frac{mg}{k}(\frac{dy}{dx}) = 0$

Using an online solver we find:

$y = -\frac{1}{2}[(1 - 4x^2)^\frac{1}{2} - ln(1 + (1 - 4x^2)^\frac{1}{2})]$

As a graph:

So, a ball rolling in this bowl (when confined between -1/2 and 1/2) will roll with an x-component in simple harmonic motion!

Neat-o!

Let's Get This Down On Paper

Sunday, September 6, 2015

Simple Harmonic Ball in Bowl?